Question

In a carnival game, a person wagers $2 on the roll of two dice. If the total of the two dice is 2, 3, 4, 5, or 6 then the person gets $4 (the $2 wager and $2 winnings). If the total of the two dice is 8, 9, 10, 11, or 12 then the person gets nothing (loses $2). If the total of the two dice is 7, the person gets $0.75 back (loses $0.25). What is the expected value of playing the game once?

Answer 1

Answer:

The expected value of playing the game once is -$0.04

Step-by-step explanation:

Consider the provided information.

If the total of the two dice is 2, 3, 4, 5, or 6 then the person gets $4 (the $2 wager and $2 winnings).

The total number of outcomes are:

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

The probability of getting sum 2 is $\frac{1}{36}$

The probability of getting sum 3 is $\frac{2}{36}$

The probability of getting sum 4 is $\frac{3}{36}$

The probability of getting sum 5 is $\frac{4}{36}$

The probability of getting sum 6 is $\frac{5}{36}$

$P(2, 3, 4, 5 or 6) = \frac{1 + 2 + 3 + 4 + 5}{36} =\frac{15}{36}$

Similarly, the probability of getting sum 8, 9, 10, 11, or 12 is:

$P(8, 9, 10, 11, or 12) = \frac{1 + 2 + 3 + 4 + 5}{36} =\frac{15}{36}$

The probability of P(7) is $\frac{6}{36}$

If the total of the two dice is 2, 3, 4, 5, or 6 then the person gets $4.

If the total of the two dice is 8, 9, 10, 11, or 12 then the person gets nothing (loses $2).

If the total of the two dice is 7, the person gets $0.75 back (loses $0.25).

Thus the required expected value is:

$\frac{15}{36}\times2-2\times\frac{15}{36}-0.25\times\frac{6}{36}$

$\frac{30}{36}-\frac{30}{36}-\frac{0.25}{6}$

$-\frac{0.25}{6}\approx-0.04$

Hence, the expected value of playing the game once is -$0.04